Question:medium
If $s = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)$ and $t = \sqrt{1 - x^2}$, then $\frac{ds}{dt}$ at $x = \frac{1}{2}$ is:
If $s = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)$ and $t = \sqrt{1 - x^2}$, then $\frac{ds}{dt}$ at $x = \frac{1}{2}$ is:
Show Hint
In parametric differentiation, if both $s$ and $t$ share a common denominator in their derivatives, they will cancel out. Focus on the numerators of $ds/dx$ and $dt/dx$ to find the final ratio quickly.
Updated On: May 2, 2026
- $1$
- $2$
- $-2$
- $4$
- $-4$
Show Solution
The Correct Option is D
Solution and Explanation
Was this answer helpful?
0
Top Questions on Derivatives of Functions in Parametric Forms
- If the lines \( \frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4} \) and \( \frac{x-3}{1} = \frac{y - 9/2}{2} = z/1 \) intersect, then the value of \( k \) is:
- MHT CET - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
- If \( 2 \tan^2 \theta - 5 \sec \theta = 1 \) has exactly 7 solutions in the interval \(\left[ 0, \frac{n\pi}{2} \right],\) for the least value of \( n \in \mathbb{N} \) then \(\sum_{k=1}^{n} \frac{k}{2^k}\) is equal to:
- JEE Main - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
The curve y(x) = ax3 + bx2 + cx + 5 touches the x-axis at the point P(–2, 0) and cuts the y-axis at the point Q, where y′ is equal to 3. Then the local maximum value of y(x) is
- JEE Main - 2022
- Mathematics
- Derivatives of Functions in Parametric Forms
- Let $x=\sin \left(2 \tan ^{-1} \alpha\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$ If $S =\left\{\alpha \in R : y ^2=1- x \right\}$, then $\displaystyle\sum_{\alpha \in S } 16 \alpha^3$ is equal to ______
- JEE Main - 2023
- Mathematics
- Derivatives of Functions in Parametric Forms
- Want to practice more? Try solving extra ecology questions todayView All Questions
